CS 332: Algorithms

David Luebke 1 04/22/23

CS 332: Algorithms

Review: Insertion Sort, Merge SortHeaps, Heapsort, and Priority Queues


Review: Merge Sort

MergeSort(A, left, right) {if (left < right) {mid = floor((left + right) / 2);MergeSort(A, left, mid);MergeSort(A, mid+1, right);Merge(A, left, mid, right);}

}

// Merge() takes two sorted subarrays of A and// merges them into a single sorted subarray of A.// Merge()takes O(n) time, n = length of A


Review: Analysis of Merge Sort

Statement Effort

So T(n) = (1) when n = 1, and 2T(n/2) + (n) when n > 1

Solving this recurrence (how?) gives T(n) = n lg n

MergeSort(A, left, right) { T(n) if (left < right) { (1) mid = floor((left + right) / 2); (1) MergeSort(A, left, mid); T(n/2) MergeSort(A, mid+1, right); T(n/2) Merge(A, left, mid, right); (n) }}


Review: Recurrences

Recurrence: an equation that describes a function in terms of its value on smaller functions

00

)1(0

)(nn

nscns

0)1(00

)(nnsnn

ns

1

22

1

)(ncnT

nc

nT

1

1)(

ncnbnaT

ncnT


Review: Solving Recurrences

Substitution method Iteration method Master method


Review: Substitution Method

Substitution Method: Guess the form of the answer, then use induction

to find the constants and show that solution works Example:

T(n) = 2T(n/2) + (n) T(n) = (n lg n) T(n) = 2T(n/2 + n ???


Review: Substitution Method

Substitution Method: Guess the form of the answer, then use induction

to find the constants and show that solution works Examples:

T(n) = 2T(n/2) + (n) T(n) = (n lg n) T(n) = 2T(n/2) + n T(n) = (n lg n)

We can show that this holds by induction


Substitution Method

Our goal: show that T(n) = 2T(n/2) + n = O(n lg n)

Thus, we need to show that T(n) c n lg n with an appropriate choice of c Inductive hypothesis: assume

T(n/2) c n/2 lg n/2 Substitute back into recurrence to show that

T(n) c n lg n follows, when c 1(show on board)


Review: Iteration Method

Iteration method: Expand the recurrence k times Work some algebra to express as a summation Evaluate the summation


s(n) = c + s(n-1)c + c + s(n-2)2c + s(n-2)2c + c + s(n-3)3c + s(n-3)…kc + s(n-k) = ck + s(n-k)

0)1(00

)(nnscn

nsReview:


So far for n >= k we have s(n) = ck + s(n-k)

What if k = n? s(n) = cn + s(0) = cn

0)1(00

)(nnscn

nsReview:


T(n) = 2T(n/2) + c2(2T(n/2/2) + c) + c22T(n/22) + 2c + c22(2T(n/22/2) + c) + 3c23T(n/23) + 4c + 3c23T(n/23) + 7c23(2T(n/23/2) + c) + 7c24T(n/24) + 15c…2kT(n/2k) + (2k - 1)c

1

22

1)( ncnT

ncnTReview:


So far for n > 2k we have T(n) = 2kT(n/2k) + (2k - 1)c

What if k = lg n? T(n) = 2lg n T(n/2lg n) + (2lg n - 1)c

= n T(n/n) + (n - 1)c= n T(1) + (n-1)c= nc + (n-1)c = (2n - 1)c

1

22

1)( ncnT

ncnTReview:


Review: The Master Theorem

Given: a divide and conquer algorithm An algorithm that divides the problem of size n

into a subproblems, each of size n/b Let the cost of each stage (i.e., the work to divide

the problem + combine solved subproblems) be described by the function f(n)

Then, the Master Theorem gives us a cookbook for the algorithm’s running time:


Review: The Master Theorem

if T(n) = aT(n/b) + f(n) then

10

largefor )()/( AND )(

)(

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)(

log)(

log

log

log

log

log

c

nncfbnafnnf

nnf

nOnf

nf

nn

n

nT

a

a

a

a

a

b

b

b

b

b


Sorting Revisited

So far we’ve talked about two algorithms to sort an array of numbers What is the advantage of merge sort? What is the advantage of insertion sort?

Next on the agenda: Heapsort Combines advantages of both previous algorithms


Heaps

A heap can be seen as a complete binary tree:

What makes a binary tree complete? Is the example above complete?


Heaps

In practice, heaps are usually implemented as arrays:

16

14 10

8 7 9 3

2 4 1

16 14 10 8 7 9 3 2 4 1A = =


Heaps

To represent a complete binary tree as an array: The root node is A[1] Node i is A[i] The parent of node i is A[i/2] (note: integer divide) The left child of node i is A[2i] The right child of node i is A[2i + 1]

16

14 10

8 7 9 3

2 4 1

16 14 10 8 7 9 3 2 4 1A = =


Referencing Heap Elements

So…Parent(i) { return i/2; }Left(i) { return 2*i; }right(i) { return 2*i + 1; }

An aside: How would you implement this most efficiently?

Another aside: Really?


The Heap Property

Heaps also satisfy the heap property:A[Parent(i)] A[i] for all nodes i > 1 In other words, the value of a node is at most the value

of its parent Where is the largest element in a heap stored?

Definitions: The height of a node in the tree = the number of edges

on the longest downward path to a leaf The height of a tree = the height of its root


Heap Height

What is the height of an n-element heap? Why?

This is nice: basic heap operations take at most time proportional to the height of the heap


Heap Operations: Heapify()

Heapify(): maintain the heap property Given: a node i in the heap with children l and r Given: two subtrees rooted at l and r, assumed to be

heaps Problem: The subtree rooted at i may violate the heap

property (How?) Action: let the value of the parent node “float down” so

subtree at i satisfies the heap property What do you suppose will be the basic operation between i, l,

and r?


Heap Operations: Heapify()

Heapify(A, i){

l = Left(i); r = Right(i);if (l <= heap_size(A) && A[l] > A[i]) largest = l;elselargest = i;if (r <= heap_size(A) && A[r] > A[largest])largest = r;if (largest != i) Swap(A, i, largest);Heapify(A, largest);

}


Heapify() Example

16

4 10

14 7 9 3

2 8 1

16 4 10 14 7 9 3 2 8 1A =


Heapify() Example

16

4 10

14 7 9 3

2 8 1

16 10 14 7 9 3 2 8 1A = 4


Heapify() Example

16

4 10

14 7 9 3

2 8 1

16 10 7 9 3 2 8 1A = 4 14


Heapify() Example

16

14 10

4 7 9 3

2 8 1

16 14 10 4 7 9 3 2 8 1A =


Heapify() Example

16

14 10

4 7 9 3

2 8 1

16 14 10 7 9 3 2 8 1A = 4


Heapify() Example

16

14 10

4 7 9 3

2 8 1

16 14 10 7 9 3 2 1A = 4 8


Heapify() Example

16

14 10

8 7 9 3

2 4 1

16 14 10 8 7 9 3 2 4 1A =


Heapify() Example

16

14 10

8 7 9 3

2 4 1

16 14 10 8 7 9 3 2 1A = 4


Heapify() Example

16

14 10

8 7 9 3

2 4 1

16 14 10 8 7 9 3 2 4 1A =


Analyzing Heapify(): Informal

Aside from the recursive call, what is the running time of Heapify()?

How many times can Heapify() recursively call itself?

What is the worst-case running time of Heapify() on a heap of size n?


Analyzing Heapify(): Formal

Fixing up relationships between i, l, and r takes (1) time

If the heap at i has n elements, how many elements can the subtrees at l or r have? Draw it

Answer: 2n/3 (worst case: bottom row 1/2 full) So time taken by Heapify() is given by

T(n) T(2n/3) + (1)


Analyzing Heapify(): Formal

So we have T(n) T(2n/3) + (1)

By case 2 of the Master Theorem,T(n) = O(lg n)

Thus, Heapify() takes linear time


Heap Operations: BuildHeap()

We can build a heap in a bottom-up manner by running Heapify() on successive subarrays Fact: for array of length n, all elements in range

A[n/2 + 1 .. n] are heaps (Why?) So:

Walk backwards through the array from n/2 to 1, calling Heapify() on each node.

Order of processing guarantees that the children of node i are heaps when i is processed


BuildHeap()

// given an unsorted array A, make A a heapBuildHeap(A){heap_size(A) = length(A);for (i = length[A]/2 downto 1)

Heapify(A, i);}


BuildHeap() Example

Work through exampleA = {4, 1, 3, 2, 16, 9, 10, 14, 8, 7}

4

1 3

2 16 9 10

14 8 7


Analyzing BuildHeap()

Each call to Heapify() takes O(lg n) time There are O(n) such calls (specifically, n/2) Thus the running time is O(n lg n)

Is this a correct asymptotic upper bound? Is this an asymptotically tight bound?

A tighter bound is O(n) How can this be? Is there a flaw in the above

reasoning?


Analyzing BuildHeap(): Tight

To Heapify() a subtree takes O(h) time where h is the height of the subtree h = O(lg m), m = # nodes in subtree The height of most subtrees is small

Fact: an n-element heap has at most n/2h+1 nodes of height h

CLR 7.3 uses this fact to prove that BuildHeap() takes O(n) time


Heapsort

Given BuildHeap(), an in-place sorting algorithm is easily constructed: Maximum element is at A[1] Discard by swapping with element at A[n]

Decrement heap_size[A] A[n] now contains correct value

Restore heap property at A[1] by calling Heapify()

Repeat, always swapping A[1] for A[heap_size(A)]


Heapsort

Heapsort(A){BuildHeap(A);for (i = length(A) downto 2){Swap(A[1], A[i]);heap_size(A) -= 1;Heapify(A, 1);}

}


Analyzing Heapsort

The call to BuildHeap() takes O(n) time Each of the n - 1 calls to Heapify() takes

O(lg n) time Thus the total time taken by HeapSort()

= O(n) + (n - 1) O(lg n)= O(n) + O(n lg n)= O(n lg n)


Priority Queues

Heapsort is a nice algorithm, but in practice Quicksort (coming up) usually wins

But the heap data structure is incredibly useful for implementing priority queues A data structure for maintaining a set S of elements,

each with an associated value or key Supports the operations Insert(), Maximum(),

and ExtractMax() What might a priority queue be useful for?


Priority Queue Operations

Insert(S, x) inserts the element x into set S Maximum(S) returns the element of S with

the maximum key ExtractMax(S) removes and returns the

element of S with the maximum key How could we implement these operations

using a heap?

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CS 332: Algorithms